If in the expansion of `(a-2b)^(n)`, the sum of `5^(th)` and `6^(th)` terms is 0, then th e values of `a//b`=

Options

A. `(n-4)/(5)`
B. `(2(n-4))/(5)`
C. `(5)/(n-4)`
D. `(5)/(2(n-4))`

Answer

B

Explanation

`T_(5) = .^(n)C_(4)a^(n-4)(-2b)^(4)`
and `T_(6) = .^(n)C_(5)a^(n-5)(-2b)^(5)`
As `T_(5) + T_(6) = 0`, we get
`.^(n)C_(4)2^(4)a^(n-4)b^(4)=.^(n)C_(5)2^(5)a^(n-5)b^(5)`
or `(a^(n-4)b^(4))/(a^(n-5)b^(5)) = (n!2^(5))/(5!(n-5)!) .(4!(n-4)!)/(n!2^(4))`
or `(a)/(b) = (2(n-4))/(5)`